2
$\begingroup$

When you graph the function (x^4)-2, the graph crosses the x axis at two points and forms one big flat trough. Since the graph is of degree 4, and has two (symmetric) roots(zeros) I would have assumed that each zero is of multiplicity 2. But I thought that a graph does not cross the x-axis at even-multiplicity zeros, but in this case, they do?

Could somebody please help me out and provide some surefire general rules for cases like this. Thanks in advance.

$\endgroup$
  • $\begingroup$ The other two roots are complex. So each root which you are talking of is of multiplicity 1. $\endgroup$ – Shailesh Oct 10 '15 at 12:33
0
$\begingroup$

If you factor this polynomial completely, you'll see that $$x^4 - 2 = (x^2 + \sqrt{2})(x^2 - \sqrt{2}) = (x + \sqrt[4]{2}i)(x - \sqrt[4]{2}i)(x + \sqrt[4]{2})(x - \sqrt[4]{2}).$$ Over $\mathbb{R}$, we have to stop at $$x^4 - 2 = (x^2 + \sqrt{2})(x^2 - \sqrt{2}) = (x^2 + \sqrt{2})(x + \sqrt[4]{2})(x - \sqrt[4]{2}),$$ because the first quadratic factor only factors over the complex numbers. This polynomial has only 2 real roots, each of multiplicity 1.

Over $\mathbb{C}$, every polynomial of degree $k$ has exactly $k$ roots, counting multiplicity (this is known as the fundamental theorem of algebra).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.